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3 years ago

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If a ski lift raises 150 passengers averaging 590 N in weight to a height of 244 m in 96.3 s, at constant speed, what average po

wer is required of the force making the lift

2 answers:

Lostsunrise [7]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

SVEN [57.7K]3 years ago

3 0

Answer:

224236.76 W

Explanation:

Power: This can be defined as the rate of change of energy. The S.I unit of power is Watt(W).

From the question,

P = E/t ......................... Equation 1

Where P = power, E = Energy or work, t = time.

But,

E = Wt×d.................. Equation 2

Where Wt = Total weight of the passengers, d = distance.

Substitute equation 2 into equation 1

P = (Wt×d)/t.............. Equation 3

Given: Wt = 590×150 =88500 N, d = 244 m, t = 96.3 s.

Substitute into equation 3

P = (88500×244)/96.3

P = 224236.76 W

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Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

$t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h$

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

$d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km$

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

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3 years ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

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Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

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3 years ago

If a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please

kherson [118]

Answer:

a = 3.125 [m/s^2]

Explanation:

In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.

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The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.

25 = 0 + a*8

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3 years ago